# Maximize distinct elements by incrementing/decrementing an element or keeping it same

Given an array arr[] of N elements, the task is to maximize the count of distinct elements in the array, by either of the given operation on each element of the array:

• either increasing the element by 1
• or decreasing the element by 1
• or keeping the element as it is.

Note: No element can be less than or equal to 0.

Examples:

Input: arr = [4, 4, 5, 5, 5, 5, 6, 6]
Output:
Explanation: After modification of each element of the array in any of the three possible ways, arr[] = [3, 4, 5, 5, 5, 5, 6, 7]. Here distinct elements are 5.

Input: arr = [1, 1, 1, 8, 8, 8, 9, 9]
Output:
Explanation: After modification of each element of the array in any of the three possible ways, arr[] = [1, 1, 2, 7, 8, 8, 9, 10]. Here distinct elements are 6.

Approach: The idea is to sort the given array first, so that the elements can be checked easily, if it is distinct, by comparing with adjacent elements.

1. First, sort all elements of the array.
2. Initialize variables count and prev to 0. (To store the count of distinct elements and previous element respectively.)
3. After that keep a track of the previous element using prev variable.
4. Iterate the sorted array.
5. Decrease the current element’s value by 1 and check if the previous element is lesser than the decreased value. If it is lesser then increment the count and assign current value to prev.
6. If the decreased value of the current element is not greater than the previous element then keep the current element as it is and check if the previous element is lesser than the current element. If it is lesser then increment the count and assign current value to prev.
7. If the current value is not greater than the previous element then increment the current value by 1 and check if the previous element is lesser than the incremented current element. If it is lesser then increment the count and assign current value to prev.
8. If incremented value of current element is not lesser than previous value then skip that element.

Below is the implementation of the above approach:

## C++

 // C++ program to Maximize distinct// elements by incrementing/decrementing// an element or keeping it same #include using namespace std; // Function that Maximize// the count of distinct// elementint max_dist_ele(int arr[],                 int n){     // sort the array    sort(arr, arr + n);     int ans = 0;     // keeping track of    // previous change    int prev = 0;     for (int i = 0;         i < n; i++) {         // check the        // decremented value        if (prev < (arr[i] - 1)) {             ans++;            prev = arr[i] - 1;        }         // check the current        // value        else if (prev < (arr[i])) {             ans++;            prev = arr[i];        }         // check the        // incremented value        else if (prev < (arr[i] + 1)) {             ans++;            prev = arr[i] + 1;        }    }    return ans;} // Driver Codeint main(){    int arr[] = { 1, 1, 1, 8,                  8, 8, 9, 9 };    int n = sizeof(arr) / sizeof(arr[0]);    cout << max_dist_ele(arr, n)         << endl;    return 0;}

## Java

 // Java program to Maximize// the count of distinct element import java.util.*; public class GFG {     // Function that Maximize    // the count of distinct element    static int max_dist_ele(        int arr[], int n)    {        // sort the array        Arrays.sort(arr);         int ans = 0;         // keeping track of        // previous change        int prev = 0;         for (int i = 0;             i < n; i++) {             // decrement is possible            if (prev < (arr[i] - 1)) {                 ans++;                prev = arr[i] - 1;            }             // remain as it is            else if (prev < (arr[i])) {                 ans++;                prev = arr[i];            }            // increment is possible            else if (prev < (arr[i] + 1)) {                ans++;                prev = arr[i] + 1;            }        }         return ans;    }     // Driver Code    public static void main(String args[])    {        int arr[] = { 1, 1, 1, 8,                      8, 8, 9, 9 };        int n = arr.length;         System.out.println(max_dist_ele(arr, n));    }}

## Python3

 # Python3 program to Maximize # the count of distinct elementdef max_dist_ele(arr, n):         # Sort the list    arr.sort()         ans = 0         # Keeping track of     # previous change    prev = 0         for i in range(n):                 # Decrement is possible        if prev < (arr[i] - 1):            ans += 1;            prev = arr[i] - 1                     # Remain as it is        elif prev < (arr[i]):            ans += 1            prev = arr[i]                     # Increment is possible        elif prev < (arr[i] + 1):            ans += 1            prev = arr[i] + 1         return ans # Driver Codearr = [ 1, 1, 1, 8, 8, 8, 9, 9 ]n = len(arr) print(max_dist_ele(arr, n)) # This code is contributed by rutvik_56

## C#

 // C# program to maximize the // count of distinct elementusing System; class GFG{ // Function that maximize the // count of distinct elementstatic int max_dist_ele(int []arr, int n){         // Sort the array    Array.Sort(arr);     int ans = 0;     // Keeping track of    // previous change    int prev = 0;     for(int i = 0; i < n; i++)    {               // Decrement is possible       if (prev < (arr[i] - 1))       {           ans++;           prev = arr[i] - 1;       }               // Remain as it is       else if (prev < (arr[i]))       {           ans++;           prev = arr[i];       }               // Increment is possible       else if (prev < (arr[i] + 1))       {           ans++;           prev = arr[i] + 1;       }    }    return ans;} // Driver Codepublic static void Main(String []args){    int []arr = { 1, 1, 1, 8,                  8, 8, 9, 9 };    int n = arr.Length;     Console.WriteLine(max_dist_ele(arr, n));}} // This code is contributed by Amit Katiyar

## Javascript



Output:
6

Time Complexity: O(N*logN)

Auxiliary Space: O(1) as it is using constant space for variables

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